What is the known vector |a|=2|, b|=3|a-2b|=5|a+2b|?

What is the known vector |a|=2|, b|=3|a-2b|=5|a+2b|?

This problem is equivalent to solving a triangle. As shown in the figure below, the sine or cosine value of all angles can be obtained by knowing the length of three sides of a triangle. For example, cosA=[(2b)^2+5^2-a^2]/2*(2b)*5, and then unknown sides can be obtained by using the cosine formula.

This problem is equivalent to solving a triangle. As shown in the figure below, the sine or cosine value of all angles can be obtained by knowing the length of three sides of the triangle. For example, cosA=[(2b)^2+5^2-a^2]/2*(2b)*5, and then unknown sides can be obtained by using the cosine formula.

Given vector a=(-1,3), B=(-2,1), find 4(2a-3b)-5(a-2b) Given vector a=(-1,3), B=(-2,1), Find 4(2a-3b)-5(a-2b)

(1,7)

Known vector A =(3,-1), B=(-1,2), then -3 A-2 The coordinates of b are ___.

By vector

A =(3,-1),

B=(-1,2),
Z-3

A-2

B=-3(3,-1)-2(-1,2)=(-9,3)-(-2,4)=(-7,-1).
Therefore, the answer is:(-7,-1).

Vector a=(root number 3,1), b=(0,-2), then a+2b collinear vector can be

A +2 b =(root 3,-3)
His collinear vector is:(x*root 3, x*-3) x is not equal to 0

In the space rectangular coordinate system OXYZ, the normal vector of the plane OAB n =(2,-2,1), given the point P (-1,3,2), then the distance from the point P to the plane is? Online equal speed! In the space rectangular coordinate system OXYZ, the normal vector of the plane OAB n=(2,-2,1), if the point P (-1,3,2) is known, then the distance from the point P to the plane is? Online equal speed!

The plane crosses the origin, so the plane equation must be in the form of Ax+By+Cz=0, and the normal vector of the plane is known, so it must be:
A =2, B =-2, C =1, because for any point (x, y, z) in the plane, the inner product of the normal vector and that point is exactly:
(X, y, z) n=2x-2y+z=0, corresponding to the equation.
Then use the point-to-plane distance formula (sqrt is the root and ^2 is the square):
D =| Ax0+ By0+ Cz0|/sqrt (A^2+ B^2+ C^2)
=|-2-6+2|/Sqrt (4+4+1)=2.
Answer:2

The vector n=(2,0,1) is the normal vector of α, a (-1,2,1) is in α, P (1,2-2) is the distance from α.

There is a formula to multiply aP vector x|x by n|/|n|aP vector by the absolute value of n vector divided by the absolute value of n vector. Sorry, no sign.

There is a set of formulas a P vector x|x by n|/|n|a P vector by the absolute value of n vector divided by the absolute value of n vector.

There is a set of formulas aP vector x|x by n|/|n|aP vector by the absolute value of n vector divided by the absolute value of n vector. Sorry, no sign.